NCEA Level 2The derivativeAS 91262

The derivative: 7 exam questions solved step by step

What this lesson covers

  • The derivative gives the gradient of the curve at every point.
  • The gradient at a point is the gradient of the tangent there, which the chord gradients approach.
Key idea

7 questions, one idea

  • 1. dy/dx is the gradient at every x.
  • 2. xⁿ → nxⁿ⁻¹, one term at a time.
  • 3. Gradient: substitute into dy/dx.
  • 4. Point on the curve: substitute into y.
Your turn

Question 1 · Achieved

  • Differentiate
  • y = 2x⁴ − 3x² + 5x − 8
Worked example

Q1: term by term

  • Power down, one less.
  • A constant goes to 0.
  1. y = 2x⁴ − 3x² + 5x − 8
  2. 2x⁴ → 4 × 2x³ = 8x³xⁿ → nxⁿ⁻¹
  3. −3x² → 2 × −3x = −6xpower 2 down
  4. 5x → 5x → 1
  5. −8 → 0a constant
  6. dy/dx = 8x³ − 6x + 5
Your turn

Question 2 · Achieved

  • Find the gradient of the curve
  • y = x³ − 4x at x = 3.
  • Careful: which do you substitute into?
Worked example

Q2: the gradient at a point

  • Differentiate first.
  • Then x = 3 into dy/dx.
  1. y = x³ − 4x
  2. x³ → 3x²power down, one less
  3. −4x → −4x → 1
  4. dy/dx = 3x² − 4
  5. x = 3: 3(3)² − 4into dy/dx, not y
  6. = 27 − 4 = 23
  7. y = 27 − 12 = 15that is the height
  8. The gradient at x = 3 is 23.
Your turn

Question 3 · Merit

  • For y = x² − 3x, estimate the gradient
  • at x = 4 using chords to x = 4.1
  • and x = 4.01. Then check with dy/dx.
Worked example

Q3: chords close in

  • Rise ÷ run for each chord.
  • Move the point closer.
  1. x = 4: y = 16 − 12 = 4
  2. x = 4.1: y = 16.81 − 12.3 = 4.51
  3. gradient = (4.51 − 4) ÷ 0.1 = 5.1
  4. x = 4.01: y = 16.0801 − 12.03
  5. = 4.0501
  6. gradient = (4.0501 − 4) ÷ 0.01
  7. = 0.0501 ÷ 0.01 = 5.01
  8. estimate: gradient ≈ 5
  9. dy/dx = 2x − 3 = 2(4) − 3 = 5
Your turn

Question 4 · Merit

  • For y = x² − 6x + 5, find the gradient
  • at x = 1, x = 3 and x = 5.
  • Describe the curve at each point.
Worked example

Q4: read the sign

  • + up, − down, 0 flat.
  • Size: how steep.
  1. y = x² − 6x + 5
  2. dy/dx = 2x − 6x² → 2x, −6x → −6, 5 → 0
  3. x = 1: 2(1) − 6 = −4negative: going down
  4. x = 3: 2(3) − 6 = 0zero: flat
  5. x = 3: y = 9 − 18 + 5 = −4the lowest point
  6. x = 5: 2(5) − 6 = 4positive: going up
  7. down, flat at (3, −4), then up
Your turn

Question 5 · Merit, in context

  • For 30 minutes after it leaves the oven,
  • a pie is at T = 0.05t² − 3t + 70 °C.
  • a) Find dT/dt at t = 10. What does it mean?
  • b) When is it cooling at 1 °C per minute?
Worked example

Q5: a rate of change

  • Units: °C per minute.
  • Cooling means negative.
  1. dT/dt = 0.1t − 30.05t² → 0.1t
  2. t = 10: 0.1(10) − 3 = −2
  3. −2 °C per minutenegative: cooling
  4. At 10 min, cooling 2 °C per min
  5. cooling 1 °C/min: dT/dt = −1
  6. 0.1t − 3 = −1
  7. 0.1t = 2+3 both sides
  8. t = 20÷ 0.1
  9. After 20 minutes
Your turn

Question 6 · Excellence

  • The curve y = ax² + bx has gradient 7
  • at x = 1, and gradient −1 at x = −1.
  • Find a and b.
Worked example

Q6: two gradients, two equations

  • a and b are numbers.
  • Each gradient: an equation.
  1. dy/dx = 2ax + bax² → 2ax, bx → b
  2. x = 1: 2a + b = 7 (1)
  3. x = −1: −2a + b = −1 (2)
  4. 2b = 6(1) + (2): a cancels
  5. b = 3÷ 2
  6. 2a + 3 = 7, so a = 2into (1)
  7. y = 2x² + 3x
  8. dy/dx = 4x + 3: 4 + 3 = 7
  9. −4 + 3 = −1
Your turn

Question 7 · Excellence

  • For the curve y = x³ + 3x − 2:
  • a) Show the gradient is always positive.
  • b) Find the points where the gradient is 15.
Worked example

Q7a: always positive

  • Reason about every x.
  • A square is never negative.
  1. dy/dx = 3x² + 3x³ → 3x², 3x → 3
  2. x² ≥ 0 for every xa square is never negative
  3. 3x² ≥ 0
  4. 3x² + 3 ≥ 3 > 0
  5. least gradient 3, at x = 0
  6. The gradient is always positive.
Worked example

Q7b: where the gradient is 15

  • Two square roots.
  • y from the curve.
  1. dy/dx = 3x² + 3
  2. 3x² + 3 = 15gradient 15
  3. 3x² = 12−3 both sides
  4. x² = 4÷ 3
  5. x = 2 or x = −2±√4
  6. x = 2: y = 8 + 6 − 2 = 12
  7. x = −2: y = −8 − 6 − 2 = −16
  8. (2, 12) and (−2, −16)
Summary

The method that solved all seven

  • Differentiate, substitute into the right one, interpret.
StepHow
Rulexⁿ → nxⁿ⁻¹, a constant → 0
Chordsrise ÷ run, closing in
Gradientsubstitute into dy/dx
Pointsubstitute into y
Sign+ up, − down, 0 flat
Contexta rate: units per unit

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